Find the Function Rule table[[x,q(x)],[1,34],[2,56],[3,78],[4,100]]

$\begin{array}{c} x & q (x) \\ 1 & 34 \\ 2 & 56 \\ 3 & 78 \\ 4 & 100 \end{array}$

Step 1

Check if the function rule is linear.

To find if the table follows a function rule, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Build a set of equations from the table such that $q (x) = a x + b$ .

$34 = a (1) + b 56 = a (2) + b 78 = a (3) + b 100 = a (4) + b$

Calculate the values of $a$ and $b$ .

Solve for $a$ in $34 = a + b$ .

Rewrite the equation as $a + b = 34$ .

$a + b = 34$

$56 = a (2) + b$

$78 = a (3) + b$

$100 = a (4) + b$

Subtract $b$ from both sides of the equation.

$a = 34 - b$

$56 = a (2) + b$

$78 = a (3) + b$

$100 = a (4) + b$

$a = 34 - b$

$56 = a (2) + b$

$78 = a (3) + b$

$100 = a (4) + b$

Replace all occurrences of $a$ with $34 - b$ in each equation.

Replace all occurrences of $a$ in $56 = a (2) + b$ with $34 - b$ .

$56 = (34 - b) (2) + b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

Simplify the right side.

Simplify $(34 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$56 = 34 \cdot 2 - b \cdot 2 + b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

Multiply $34$ by $2$ .

$56 = 68 - b \cdot 2 + b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

Multiply $2$ by $- 1$ .

$56 = 68 - 2 b + b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

$56 = 68 - 2 b + b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

Add $- 2 b$ and $b$ .

$56 = 68 - b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

$56 = 68 - b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

$56 = 68 - b$

$a = 34 - b$

$78 = a (3) + b$

$100 = a (4) + b$

Replace all occurrences of $a$ in $78 = a (3) + b$ with $34 - b$ .

$78 = (34 - b) (3) + b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

Simplify the right side.

Simplify $(34 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$78 = 34 \cdot 3 - b \cdot 3 + b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

Multiply $34$ by $3$ .

$78 = 102 - b \cdot 3 + b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

Multiply $3$ by $- 1$ .

$78 = 102 - 3 b + b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

$78 = 102 - 3 b + b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

Add $- 3 b$ and $b$ .

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = a (4) + b$

Replace all occurrences of $a$ in $100 = a (4) + b$ with $34 - b$ .

$100 = (34 - b) (4) + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Simplify the right side.

Simplify $(34 - b) (4) + b$ .

Simplify each term.

Apply the distributive property.

$100 = 34 \cdot 4 - b \cdot 4 + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Multiply $34$ by $4$ .

$100 = 136 - b \cdot 4 + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Multiply $4$ by $- 1$ .

$100 = 136 - 4 b + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = 136 - 4 b + b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Add $- 4 b$ and $b$ .

$100 = 136 - 3 b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = 136 - 3 b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = 136 - 3 b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$100 = 136 - 3 b$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Solve for $b$ in $100 = 136 - 3 b$ .

Rewrite the equation as $136 - 3 b = 100$ .

$136 - 3 b = 100$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Move all terms not containing $b$ to the right side of the equation.

Subtract $136$ from both sides of the equation.

$- 3 b = 100 - 136$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Subtract $136$ from $100$ .

$- 3 b = - 36$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$- 3 b = - 36$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Divide each term in $- 3 b = - 36$ by $- 3$ and simplify.

Divide each term in $- 3 b = - 36$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{- 36}{- 3}$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{- 36}{- 3}$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Divide $b$ by $1$ .

$b = \frac{- 36}{- 3}$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$b = \frac{- 36}{- 3}$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$b = \frac{- 36}{- 3}$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Simplify the right side.

Divide $- 36$ by $- 3$ .

$b = 12$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$b = 12$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$b = 12$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

$b = 12$

$78 = 102 - 2 b$

$56 = 68 - b$

$a = 34 - b$

Replace all occurrences of $b$ with $12$ in each equation.

Replace all occurrences of $b$ in $78 = 102 - 2 b$ with $12$ .

$78 = 102 - 2 \cdot 12$

$b = 12$

$56 = 68 - b$

$a = 34 - b$

Simplify the right side.

Simplify $102 - 2 \cdot 12$ .

Multiply $- 2$ by $12$ .

$78 = 102 - 24$

$b = 12$

$56 = 68 - b$

$a = 34 - b$

Subtract $24$ from $102$ .

$78 = 78$

$b = 12$

$56 = 68 - b$

$a = 34 - b$

$78 = 78$

$b = 12$

$56 = 68 - b$

$a = 34 - b$

$78 = 78$

$b = 12$

$56 = 68 - b$

$a = 34 - b$

Replace all occurrences of $b$ in $56 = 68 - b$ with $12$ .

$56 = 68 - (12)$

$78 = 78$

$b = 12$

$a = 34 - b$

Simplify the right side.

Simplify $68 - (12)$ .

Multiply $- 1$ by $12$ .

$56 = 68 - 12$

$78 = 78$

$b = 12$

$a = 34 - b$

Subtract $12$ from $68$ .

$56 = 56$

$78 = 78$

$b = 12$

$a = 34 - b$

$56 = 56$

$78 = 78$

$b = 12$

$a = 34 - b$

$56 = 56$

$78 = 78$

$b = 12$

$a = 34 - b$

Replace all occurrences of $b$ in $a = 34 - b$ with $12$ .

$a = 34 - (12)$

$56 = 56$

$78 = 78$

$b = 12$

Simplify the right side.

Simplify $34 - (12)$ .

Multiply $- 1$ by $12$ .

$a = 34 - 12$

$56 = 56$

$78 = 78$

$b = 12$

Subtract $12$ from $34$ .

$a = 22$

$56 = 56$

$78 = 78$

$b = 12$

$a = 22$

$56 = 56$

$78 = 78$

$b = 12$

$a = 22$

$56 = 56$

$78 = 78$

$b = 12$

$a = 22$

$56 = 56$

$78 = 78$

$b = 12$

Remove any equations from the system that are always true.

$a = 22$

$b = 12$

List all of the solutions.

$a = 22, b = 12$

Calculate the value of $y$ using each $x$ value in the relation and compare this value to the given $q (x)$ value in the relation.

Calculate the value of $y$ when $a = 22$ , $b = 12$ , and $x = 1$ .

Multiply $22$ by $1$ .

$y = 22 + 12$

Add $22$ and $12$ .

$y = 34$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 34$ and $q (x) = 34$ .

$34 = 34$

Calculate the value of $y$ when $a = 22$ , $b = 12$ , and $x = 2$ .

Multiply $22$ by $2$ .

$y = 44 + 12$

Add $44$ and $12$ .

$y = 56$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 56$ and $q (x) = 56$ .

$56 = 56$

Calculate the value of $y$ when $a = 22$ , $b = 12$ , and $x = 3$ .

Multiply $22$ by $3$ .

$y = 66 + 12$

Add $66$ and $12$ .

$y = 78$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 78$ and $q (x) = 78$ .

$78 = 78$

Calculate the value of $y$ when $a = 22$ , $b = 12$ , and $x = 4$ .

Multiply $22$ by $4$ .

$y = 88 + 12$

Add $88$ and $12$ .

$y = 100$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 100$ and $q (x) = 100$ .

$100 = 100$

Since $y = q (x)$ for the corresponding $x$ values, the function is linear.

The function is linear

Step 2

Since all $y = q (x)$ , the function is linear and follows the form $y = 22 x + 12$ .

$y = 22 x + 12$

Find the Function Rule table[[x,q(x)],[1,34],[2,56],[3,78],[4,100]]

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